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EXPLANATION OF FLUORINE IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 9, 2015 Fluorine is a chemical element with symbol F and atomic number 9. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light (FASTER THAN LIGHT), which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Fluorine including the following ground state electron configuration: 1s2.2s2.2px2.2py2.2pz1. According to the “Ionization energies of the elements-WIKIPEDIA” we observe ( in eV) the following energies: E1 = 17.42282, E2 = 34.97082 , E3 = 62.7084 , E4 = 87.1398 , E5 = 114.2428 , E6 = 157.1651 . E7 = 185.186 , E8 = 953.9112, and E9 = 1103.1176 . ( See also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS). For simplicity we start with the E5 , E4 , and E3 , while the E2 and E1 are explained in the second chapter. EXPLANATION OF E5 = 114.2428 eV, E4 = 87.1398 eV, AND E3 = 62.7084 eV The 2px2 , 2py2 , and 2pz1 orbitals of charge (- 5e) under a perfect screening of the spherical shells 1s2 and 2s2, should lead to the effective ζ = 5, because + 9e -2e -2e = + 5e. However the deformation of shells leads to ζ > 5 . Of course in the absence of the electrons 2px2 and 2py2 one gets the E(2pz1) = - E5 which is the binding energy of the one electron with n = 2 based on the Bohr model. In this case of n = 2 we must find the effective ζ1 > 5, while for a perfect screening (ζ = 5) due to the spherical shells of 1s2 and 2s2 we should write to find E (2pz1) = (-13.6057) ζ2/22 = (-13.6057)52/22 = - 85.0356 eV However the 2pz1 penetrates the 2s2 which leads to the deformations not only of spherical shells 1s2 and 2s2 but also of 2pz1. Thus writing E5 = 114.2428 eV = - E(2pz1) = - (- 13.6057) ζ12/22 we get ζ1 = 5.795 > 5 . Then adding the second electron 2py1 one gets the binding energy E (2pz1 + 2py1) = -( E5 + E4 ). That is E(2pz1 + 2py1) = 2(-13.6057)ζ22/22 = - (E5 + E4 ) = - ( 114.2428 + 87.1398 ) = - 201.3826 eV Therefore one gets ζ2 = 5.44 > 5 Of course for getting the binding energy of the three electrons we may write E ( 2pz1 + 2py1 + 2px1 ) = 3(-13.6057) ζ32/22 = - ( E5 + E4 + E3) = - 264.09 eV. Hence one gets ζ3 = 5 . Here ζ3 = 5 means that the three electrons of 2px1, 2py1, and 2pz1 orbitals try to be at symmetrical positions leading to a perfect screening of the spherical shells 1s2 and 2s2 . Because of parallel spin (S = 1 ) here we find three separated electrons of 2pz1 , 2py1 and 2px1 with mutual electric and magnetic repulsions: Fem = Fe + Fm . However in the absence of 1px1 the two electrons of 2py1 and 2pz1 break the symmetry and lead to ζ2 = 5.44 > 5 . Moreover in the absence of 2px1 and 2py1 the one electron of 2pz1 orbital breaks more the symmetry and leads to ζ1 = 5.79 > 5. That is, ζ1 > ζ2 > ζ3 . ' ' EXPLANATION OF E2 = 34.97082 eV AND E1 = 17.42282 eV BASED ON MY FORMULA ' Adding also the fourth and the fifth electron one gets the total binding energy of 5 electrons as E( 2px2 + 2py2 + 2pz1) = - ( E5 + E4 + E3 + E2 + E1 ) Here the binding energy of two electrons of opposite spin ( say the 2px2) is due to a vibration energy (Ev) discovered by me in 2008: Ev = (16.95ζ - 4.1)/n2 However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of the 2px2 pair, today many physicists believe that it is due to the Coulomb repulsion between the two electrons. Under such fallacious ideas I published in Ind. J. Th. Phys. (2008) my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures”. For example the binding energy of the two electrons of opposite spin E(2px2) is given by E(px2 ) = + (16.95)ζ - 4.1 / 22 Thus the total binding energy of the five electrons can be written as E( 2px2 + 2py1 + 2pz1) = 2+ (16.95)ζ - 4.1 /22 + (-13.6057)ζ2/22 = - ( Ε1 + Ε2 + Ε3 + Ε4) Since ( Ε1 + Ε2 + Ε3 + Ε4 ) = 316.48464 eV we may write 17ζ2 - 8.475ζ - 314.43464 = 0 Then solving for ζ we get ζ = 4.557 < 5. Since ζ < 5 cannot exist, we suggest that the correct ζ = 5 of a perfect screening gives n > 2. Under this condition we find that n = 2.2 . It means that the orbitals 2px2 , 2py2 , and 2pz1 do not lead to the deformation of 1s2 and 2s2 but differ from the symmetry of the (2px1 + 2py1+ 2pz1 ), because here the paired electrons of opposite spin like the 2px2 and 2py2 do not repel magnetically the 2pz1. Thus, in the absence of any magnetic repulsion we observe the same perfect screening under n = 2.2 . ' ''' '''EXPLANATION OF E6 = 157.1651 eV AND E7 = 185.186 eV For simplicity we start with the E7 = -E( 2s1). According to the quantum mechanics the one electron (2s1) penetrates the 1s2 shell. Thus it leads to the deformations of both 1s2 and 2s1 spherical shells giving an effective ζ > 7 because the charge (- 2e) of the two electrons of 1s2 screens the charge (+9e) of nucleus. Since n = 2 we may write E7 = 185.186 eV = -E(2s1) = - (-13.6057)ζ2/22 Therefore one gets ζ = 7.38 > 7 Then adding the second electron of opposite spin we get the binding energy, E(2s2), of the two electrons of opposite spin for n = 2 given by my formula of 2008: E(2s2) = + (16.95)ζ - 4.1 / 22 = - (E6 + E7 ) Since ( E6 + E7) = 342.3511 eV, we may write 6.8025ζ2 - 4.2375ζ - 341.3261 = 0 Then, solving for ζ we get ζ = 7.4 > 7 In other words we observe that the repulsions 1s2-2s1 and 1s2-2s2 give almost the same effective ζ because the two electrons (2s2) of opposite spin behave like one particle. Whereas in the case of the three electrons of parallel spin (S=1) we observe a perfect screening, because the three electrons of 2px1, 2py1, and 2pz1 ,having parallel spin (S = 1 interact at a moment with electric and magnetic repulsions from symmetrical positions. EXPLANATION OF Ε8 = 953.9112 eV AND E9 = 1103.1176 eV As in the case of the helium the ionization energy E9 = 1103.1176 eV = - E(1s1) is due to the one remaining electron of 1s1 with n = 1 given by applying the simple Bohr model for Z = 9 as E9 = - (-13.6057 )Z2/12 = - (-13.6057)9 2 /12 = 1102.062 eV. Surprisingly one sees here that after the ionizations the Bohr model gives the value of 1102.062 eV which is smaller than the experimental value of 1103.1176. Under this condition of ionizations I suggest that n = 1 becomes n < 1 due to the fact that the ionizations reduce the electron charges and now the nuclear charge is much greater than the electron charge of the remaining electron. Therefore we may write E9 = (13.6057) Z2/n2 = (13.6057)92/n2 = 1103.1176 Then solving for n we get n = 0.99952. In the same way for calculating the E8 = 953.9112 eV we must apply my formula of 2008 as E8 = 953.9112 eV = - E9 - E(1s2) = - 1103.1176 - [ (-27.21)92 + (16.95) 9 - 4.1]/n2 Then solving for n we get n = 0.99964. Here 0.99964 > 99952 because the second electron increase the electron charge with respect to nuclear charge. Category:Fundamental physics concepts